Sorry for the reupload. I actually had a mistake on the integral of 1/(x^2-a^2) in the previous video. I will make up to you guys by checking my answer by differentiation! That video will be done soon!

you know something is hard when blackpenredpen uses five colours

Imagine doing all this and then forgetting the +c

"1+1 is 2, right?" Calculus, everybody.

I think my professor summed up integration in a nice way. He said differentiation is all about technique. You see a scenario and have a set of rules you then follow. Integration is a form of art. It's much more intricate and delicate.

"Welcome to the Salty Spitoon, how tough are ya?" "How tough am I? I just integrated a trig function!" "Yeah, so?" "integral(sqrt(tan x))dx" "Uh, right this way..."

Mr Math Man. Math Me a Man. Make him integrate the square root of tan.

Me when finding this channel: "wtf is going on?!" Me rewatching 1 year later after having seen every bprp video: "alright easy didn't even need the DI setup"

this is pure game. There are very very few maths teachers at level of you. Thank you.

Wow..you demonstrated this before 100k subs more than 3 years ago, today you have 687k. Wishing you for the next 313k

I'm preparing a transfer exam for Korean universities and there was this question on my preparation problem set. Your solution was so helpful brother, thanks a lot!

Integration is just the easiest thing ever... I can integrate √tanx + e^x² in seconds: Set up the integral: ∫ √(tan(x)) + e^x² dt And then just use the "inverse" power rule: (t)√(tan(x))+ (t)e^x² And we're done... I didn't say that I'ld do it with respect to x...

Evil integral to place on an exam ... :/

god damn I love math

Very well work dude, the resolution was easier than I thought, I had problems when using trinomio. You got a like and a new subscriber!

Ok, so in the beginning we have a pretty simple mathematical expression while the result in the end is something horrible. Things tend to evolve naturally from more complex to simpler. I therefore consider it normal to leave this formula unintegrated. Thank you for all your thumbs up ! :D

I passed my semester of Calc 1! I did not fully understand everything but I believe I got a strong majority of it & I will be working on some of my weaknesses during break to prepare for Calc 2. For some reason when we got to U-substitution, everyone was confused but it seemed to make sense to me just based off the few example she gave in class and somehow that was all I needed. Anyway, so during last class before our final, she did a little review & answered questions. Someone asked about this sqrt(tanx) and she was like "oh you cant solve this with your current toolset, wait till next semester" and I was like "I can solve it" because I had all this false confidence from having understood the whole U-sub stuff. Well turns out this is a VERY difficult one lol. I guess if you are a student and want to be prepared for integrals, just spend a week studying just THIS one integral and you should be good to go lol.

15:02 WIZARD! Where can I buy this magic blackpen??

I just saw you instagram reel, where you give credit question and searched for this integral on KZfaq. And I am so glad I found your video 😀!

14:41 "This is the two, so what should I do?" What a rhyme :O

Oh man, I love this video. I watched the entire thing and enjoyed every second of it! Keep up the good work on your channel. :)

You, my good sir, are turning calculus into art! Awesome video!

I LITERALLY LOVE YOU SO MUCH YOU DESERVE THE WHOLE WORLD

I like this integral and its anti-derivative. I am impressed with your technique of using trigonometric and u-substitution along with algebraic manipulation to arrive at the answer.

You made me fall in love with mathematics!❤️ Thank you!

When you do all of this in the exam and you realize the integral was sqrt(tanx + 1)

The most difficulty integral that i ever seen in my entire life! But it was really good xD

I majored in physics in school and always preferred the more pure abstract mathematical parts of it. Watching this video is like taking a mental vacation back into the past. I'm happy that I was able to follow it through to the end on my first viewing :)

Simple presentation of the difficult integral in a nice manner . Thanks .

Gratulálok, lenyűgöző végig az átváltások sorozata ! Főleg az (U^2 + (1/U)^2) átalakítása !

Riveting :) Bravo on a brilliant solution! I think I could explain that to someone else now!

Your reslly entertaining and it is very interesting. I love your out of the box thinking to manipulate things to make them work!,,

I just differentiated this myself, it starts out ridiculously complex, but it slowly starts to fit in with everything, good luck on doing it! You might need 6 boards to do it, I managed to do it in 1 board, but I had to rub out a lot of the work out I did

Man, so many colors! If you wrote blackpenredpen in that empty space and taken a picture, you'd have a pretty good channel banner.

10/10 what a trip

I always understand your calculus explanations. Thank you.

So... u substitute for fx Square u = sqrt (tanx). See that 2udu = sec^2(x) dx. Squaring u^2 = tanx gives us tan^2(x) = sec^2 (x) -1 = u^4 which we can see as sec^2 (x) = u^4 + 1. Thus dx = 2udu/u^4 + 1. Plug in original equation to have integral of u*(2u/u^4 + 1)du. Multiply top and bottom by 1/u^2 to get complex fraction with sum of squares in denominator. Complete the square to get (u + 1/u)^2 - 2, which has derivative of inside equal to 1 - 1/(u)^2. Now we want two integrals (why? it is not clear unless you see the tanh^-1 and tan^-1 option coming up), one with 1 - 1/(u)^2 in numerator and other with 1 + 1/(u)^2 in our numerator. Because the completed square can have two forms we can have the appropriate denominators to do two more substitutions, this time with t and say w. If we do the substitutions correctly we have two integrals, one being of 1/(t^2 - 2), and our other being of 1/(w^2 + 2), both in their respective worlds. A formula exists for these forms to be integrated neatly into tanh^-1 and tan^-1 forms. Substitute u back in for t, w, and sqrt (tanx) for u. Do this correctly, and then Add c and we're done. I did this mostly for my own understanding, but I'm fairly sure I didn't skip too much for it to act as a quick summary.

10:54 defenitely most important part

And pretend nothing happened!?? Great lines

Finally! Loved the video! Is great! You could do the same integral for the 100k, but now you do it the hard way! (The one with partial fractions and with the factorization of u^4+1, and with the natural log result!)

Ez annyira bonyolult, hogy nincs értelme ennyit számolni, de blackpenredpen igazi zseni !

Love your videos sir, you make very complex calculus part easy 😃😃

Simply great! Thank you a lot master.

Fantastic video. Very well explained. Thx

Got to see this question first time on my exam today... carrying weightage of 6marks.. was totally fucked up😑

Watching for second time, now i got it! :D Great video!!!

6:13 out of context is beautiful

Slightly more straightforward (although much longer/messier): Factor x^4 + 1 = (x^2 + sqrt(2)x+1)(x^2 - sqrt(2)x+1), then do partial fractions, complete the square in the denominators and solve. This saves you from having to figure out the trick where you add 1/u^2 and subtract 1/u^2.

you are awesome men this appear on my exam from yesterday i lov u 💕

10:43 that's your brilliancy sir

you are seriously the best maths teacher!

Brilliant. So many techniques in one shot.

Great explanation! Thank you! :)

What a gorgeous way to find the answer! Who is the genius first to find u+1/u and u-1/u pairs for this question? It really drive me crazy for such an integral. Thank BPRP.

This is lengthy problem. Very few can solve in first time. We can only solve this problem if we practice at home. Your explanation is very nice

Yesss! This question was on my calc 2 final exam and I got it right!

It feels nice that i solved it all by myself for the most part. I have some amazing teachers. Can't thank them enough

YAY I got this correct Imma do a video on how I did it and then compare it with your method. This took me ages btw

Very good explanation... Thanks a lot

sqrt(-1) just love your videos! I'm gonna start studying math very soon and your videos really hype me for it^^

Absolutely amazing 👍🏻

Great explanation!👍

Nagyon jó és elegáns megoldás !

He didn't reupload this video, we all just have a déjà-vu at the exact same moment.

This turned to blackpenredpengreenpenbluepenpurplepen real quick

Couldn’t you have just done tan inverse with the original u equation after you divided the fraction by u^2, or does it only work with one variable and one constant term?

19:05 Try to differentiate THIS to give sqrt(tan x)

the sheer joy with which my man just said: "I also have a purple pen :)"

Thank youuuu❤, greetings from 🇨🇴

Is this Calc 2?

Cool to see it done in an alternative way :)

Gracias, haces todo fácil Saludos desde Colombia

prolly the first time im actually being happy after a maths answer

I needed to do HW tonight but I watched this whole video instead. After seeing one too many memes about integrating the square root of tanx I couldn't resist.

Did you guys notice how he changed his pens at 05:23? That was awesome!

I couldn't do this integral without u

Best birthday gift , thanks

Hey man i love your vids.My question here is how can u divide with u^2 ? I have actually solved this integral with your way and a different one i got two different results. I think this way is the wrong one bcz of this.

The hardest part about this is to remember the +C.

OMG my brain is dieing... I think I'll need to watch this one more time to fully understand the integral

Every integral tackled for the first time is a journey into the unknown...

As my Cal 3 Professor used to say " What could be simpler ."

One rather interesting thing about that expression is that (if we take C = 0) we'll generally get a complex number out of it. The imaginary part is always the same though. I think you get a nice expression if you introduce complex numbers (starting by factoring (u^4 + 1) into (u^2 +- i). But it seems non-obvious that 1) the manipulations I did after that are totally ok, since I kind of ignore principle value etc... and 2) how to rigouously show this is actually equal to the answer found with real math. Writing R = (1 + i) / sqrt(2) and u = sqrt(tan(x)) (and Re == real part) then I eventually got that Re[2R*arctan(Ru)] is an antiderivative of sqrt(tan(x)) w.r.t. x Writing out the def. of (complex) arctan => Re[(1/R) Log [(R - u) / (R + u)]] is an antiderivative. Numerically checking for some random values I find this does give the same values (up to a constant) as your solution. Though I did assume that tan(x) is non-negative here (that is how I ended up with "real part" in there).

I guess I'll comment again. You could do e^(-x^2) for a 100k!

Amazing explained

Just... beautiful !

was trying to do on my own but was stuck after first substitution this video helped.

Hey bprd I have suggestion in 3:31 , we can take u^2 as t and then do partial fraction , it will be easy , after doing partial fraction substitute t as u^2 .

We can solve this in some other way too. We can write sqrt tanx as 1/2 {(sqrt tanx + sqrt cotx)+ (sqrt tanx - sqrt cotx)},and then break these into sin cos expressions,and then subtitute sinx + cosx = t and sinx-cosx = k in the first and second integrals respectively,and then apply the standard integral formula. Anyways love you process too!

You are a brilliant Math teacher

I think that the most monumental achievement humanity could ever hope to accomplish would be to find an intuitive understanding of how to go directly from the integrand to the antiderivative in one step...

This is fantastic. I am better at understanding the calculus than I am at manupulating algebra. Every time I see this kind of twsting or algebra tecniques, I wonder how the people could come up with it in a test..... By the way, have you ever used it in the final exame of your class? If so, has anyone got it right? 只是好奇啦

You could've used another formula/ method for integrating the 1/(t^2 -2) just by using : integral of 1/(x^2-a^2) = (1/2a)*( ln( |x-a|/|x+a| ) . that would have been more easy and intiutive than hyperbolic inverse function.(just saying) BTW very nice explanation sir. Great content. You're a wonderful teacher. PEACE

About the integral answer with inverse tanh and so on: is this integral defined anywhere as a function? A trip to Desmos seems to indicate that this function has no real values, and plugging in several of the usual trig values yields "undefined" no matter where I look.

Mind blown after watching this after I have forgotten mathematics...gotta study again

I passed calc 2 this summer and then I looked at this video because I couldn't figure this out on my own. I heard "Hyperbolic tan" and had a mental breakdown and screamed aloud, "SINCE BLOODY WHEN IS THERE A THING CALLED HYPERBOLIC TANGENT!? THEY'RE JUST MAKING NEW THINGS TO MESS WITH ME!" followed by expletives and crying.

Magnificent!!

I now know I got this question wrong on my final

50% math 50% flexing different colored pens

This is the most chaotic BPRP video I've ever seen